Problem: $\vec w = (2,1)$ $-6\vec w= ($
Answer: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $-6 \vec{w}$ : $\begin{aligned} {-6}\vec w = {-6} \cdot (2,1) &= \left({-6} \cdot 2, {-6} \cdot 1\right) \\\\ &= (-12,-6) \end{aligned}$ The answer is $ (-12,-6) $.